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MrRa [10]
3 years ago
5

What would decrease the resistance of wires carrying an electric current

Physics
2 answers:
Anettt [7]3 years ago
6 0
The answer is A. shorter wires

rewona [7]3 years ago
4 0
Make the wires shorter, thicker, and cooler.
Also, use silver wires. If those are too expensive, then at least make sure you're using copper wires.
You might be interested in
Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between
Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

m_1v_1+m_2v_2=(m_1+m_2)v_f

Our values are given by,

m_1=m_2=72Kg

As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

v_{x1}=5.45m/s

v_{y1}=0

Skate 2:

v_{x2} = 5.45*cos105= -1.41m/s

v_{y2} = 5.45*sin105 = 5.26m/s

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}

0+75*5.25=(75+75)v_{fy}

v_{fy}=\frac{5.25}{2}

v_{fy}=2.63m/s

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

6 0
3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h
Ratling [72]

Answer

given,

time  = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ =  difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

a_1=\dfrac{1.39}{10}

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

a_3=\dfrac{5.282}{10}

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = \dfrac{0.3892}{0.139}

ratio = 2.8

6 0
4 years ago
Why can't i tell my crush how i feel
fomenos
You can, just tell them
6 0
3 years ago
Read 2 more answers
The x component of vector is -27.3 m and the y component is +43.6 m. (a) What is the magnitude of ? (b) What is the angle betwee
lara [203]

Answer:51.44 units

Explanation:

Given

x component of vector is -27.3\hat{i}

y component of vector is 43.6\hat{j}

so position vector is

r=-27.3\hat{i}+43.6\hat{j}

Magnitude of vector is

|r|=\sqrt{27.3^2+43.6^2}

|r|=\sqrt{2646.25}

|r|=51.44 units

Direction

tan\theta =\frac{43.6}{-27.3}=-1.597

vector is in 2nd quadrant thus

180-\theta =57.94

\theta =122.06^{\circ}

4 0
3 years ago
The howler monkey is the loudest land animal and, under some circumstances, can be heard up to a distance of 8.9 km. Assume the
Airida [17]

Answer:

\frac{I}{I_0}=113.68

Explanation:

P = Acoustic power = 63 µW

r = Distance to the sound source = 210 m

Acoustic power

P=IA\\\Rightarrow I=\frac{P}{A}\\\Rightarrow I=\frac{63\times 10^{-6}}{4\times \pi \times 210^2}

Threshold intensity = I_0=1\times 10^{-12}\ W/m^2

Ratio

\frac{I}{I_0}=\frac{\frac{63\times 10^{-6}}{4\times \pi \times 210^2}}{1\times 10^{-12}}\\\Rightarrow \frac{I}{I_0}=113.68

Ratio of the acoustic intensity produced by the juvenile howler to the reference intensity is 113.68

6 0
3 years ago
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